Question

In Fig. find the horizontal velocity $u$ (in ${ms}^{-1}$) of a projectile so that it hits the inclined plane perpendicularly. Given $H=6.25m$.

Answer 1

$v_x=u_x+a_{x}t$

$\Rightarrow 0=u\cos {30}^o-g\sin {30}^{o}t$

$\Rightarrow t=\frac{u\sqrt{3}}{g}$

$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$\Rightarrow -H\cos {30}^o=-u\sin {30}^{o}t-\frac{1}{2}g\cos {30}^o{t}^2$

$\Rightarrow -H\frac{\sqrt{3}}{2}=\frac{-u}{2}\frac{u\sqrt{3}}{g}-\frac{1}{2}g\frac{\sqrt{3}}{2}\frac{u^{2}3}{g^2}$

$\Rightarrow u=\sqrt{2gH/5}=\sqrt{2\times 10\times 6.25/5}=5m/s$